$f(x, y) = x^2 + y^2 + xy$ $\dfrac{\partial^2 f}{\partial x \partial y} = $
Answer: Taking a mixed partial derivative is when we take two or more regular partial derivatives in a row, but each is with respect to a different variable. $\dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial y} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial x \partial y} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial y} \left[ x^2 + y^2 + xy \right] \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ 0 + 2y + x \right] \\ \\ &= 0 + 1 \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial x \partial y} = 1$.